Solution. Let $D=AH\cap BC$, $K=AC\cap OX$, $H'=AH\cap(ABC)\neq A$, $M$ the midpoint of $AH$, and finally $A'$ and $C'$ the $A$- and $C$-antipode in $(ABC)$, respectively. We shall prove that $K$ is the centroid in $\bigtriangleup XC'C$. Note that $XA\parallel BH\perp AC$ so $C'$, $A$, $X$ lie on the same line. Since $AXHB$ and $AHBC'$ are parallelograms, $AX=BH=AC'$ so $A$ is the midpoint of $XC'$. Note that $O$ is the midpoint of $CC'$ so the result follows. Now we note that $OH\parallel BH\parallel A'H'$ so $H$ is the midpoint of $AH'$ since $O$ is the midpoint of $AA'$. Since $H'$ is the reflection of $H$ in $D$, we must have $HD=H'D=HM=AM$. Finally we have $$\frac{AM}{AD}=\frac{1}{3}=\frac{AK}{AC}$$so $MK\parallel DC\perp AH$ which solves the problem.

Attachments:

Let $AH$ meet $BC$ at $D$, $B'$ the reflection of $X$ over $A$, $N$ the midpoint of $AH$, $M$ the midpoint of $BC$, $G$ the centroid of $\triangle ABC$ and let $OX$ meet $AB$ at $G'$. Notice that $B'$ is the B-antipode in $(ABC)$ since there is a homothety centered at $X$ sending $N$ to $C$ and $A$ to $B'$ meaning that $B'C \parallel AH$ so $AHCB'$ is a paralelogram but the only point $B'$ that holds that is exactly the B-antipode in $(ABC)$, also we have to note that $O,H,G$ lie on the euler line so $AG=2GM$ means by the parallel that $AH=2HD$ so $AN=HN=HD$ meaning that $ND=2AN$. We finally from this note that $G'$ is the centroid of $\triangle BB'X$ hence $G'B=2AG'$ meaning that $GN \parallel BC$ so $G$ lies in the perpendicular bisector of $AH$, thus we are done

Complex bash Note that $S$ be intersection of $AB$ and $XO$ ; $M$ be midpoint of $AH$. We need to show that $SM \bot AH$ Let $(ABC)$ be unit circle , so that $|a|=|b|=|c|=1$ $$h=a+b+c$$$$x+c=a+h \iff x=2a+b \wedge \bar{x} = \frac{2b+a}{ab}$$$$m=\frac{a+h}{2} \iff m=\frac{2a+b+c}{2} \wedge \bar{m}=\frac{2bc+ac+ab}{2abc}$$$$s=\frac{(\bar{a}b-a\bar{b})x}{(\bar{a}-\bar{b})x-\bar{x}(a-b)} \iff s=\frac{2a+b}{3} \wedge \bar{s}=\frac{2b+a}{3ab}$$Since $BC||OH$ $$ \frac{b-c}{h}=\frac{\bar{b}-\bar{c}}{\bar{h}} \iff \frac{b-c}{a+b+c}=\frac{a(c-b)}{ab+bc+ca} \iff a^2+2ab+2ac+bc=0(\clubsuit)$$ Now Let us finish problem $$SM \bot AH \iff \frac{h-a}{m-s}+\frac{\bar{h}-\bar{a}}{\bar{m}-\bar{s}}=0 \iff \frac{1}{b+3c+2a}+\frac{a}{2bc+ca+3ba}=0 \iff a^2+2ab+2ac+bc=0 \text { correct by } (\clubsuit)$$ so we are done